Jason Hill
Math2400 003
WeBWorK assignment number ExamReview is due : 05/09/2012 at 05:00am MDT.
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Give 4 or 5 significant digits for (floating point) numerical answers. For most problems when entering numerical answers,
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so we can add 1 to both sides to get x + 6y − z + 1 = 1. Thus,
f (x, y, z) = x + 6y − z + 1 has level surface f = 1 identical to the
graph of g(x, y) = x + 6y.
Of course, we could also say that the graph of g(x, y) is equivalent to z − x + 6y = 0, so that f = z − x − 6y + 1.
1. (0 pts) Library/CU/Michigan/Chap12Sec1/Q31v2.pg
Find the equations of planes that just touch the sphere (x −
1)2 + (y − 1)2 + (z − 5)2 = 25 and are parallel to
and
(a) The xy-plane:
(b) The yz-plane:
and
(c) The xz-plane:
and
SOLUTION
The sphere has center at (1, 1, 5) and radius 5.
(a) The planes parallel to the xy-plane just touching the
sphere are 5 above and 5 below the center. Thus, the planes
z = 10 and z = 0 are both parallel to the xy-plane and touch the
sphere at the points (1, 1, 10) and (1, 1, 0).
(b) The planes parallel to the yz-plane just touching the
sphere are 5 to the left of and 5 to the right of the center. Thus,
the planes x = 6 and x = −4 are both parallel to the yz-plane and
touch the sphere at the points (6, 1, 5) and (−4, 1, 5).
(c) The planes parallel to the xz-plane just touching the
sphere are 5 to the left of and 5 to the right of the center. Thus,
the planes y = 6 and y = −4 are both parallel to the xz-plane and
touch the sphere at the points (1, 6, 5) and (1, −4, 5).
Correct Answers:
• x+6*y-z+1
3. (0 pts) Library/CU/Michigan/Chap12Sec6/Q19-hilljb.pg
For the function f (x, y) below, determine whether there is a
value for c making the function continuous everywhere. If so,
find it.
(
c + y x ≤ 4,
f (x, y) =
6−y x > 4
c=
(If there is no value of c that works, enter none, and be sure that
you can explain why there is no such value.)
SOLUTION
The function f is continuous at all points (x, y) with x 6= 4.
So let’s analyze the continuity of f at the point (4, a). We have
•
•
•
•
•
•
z
z
x
x
y
y
=
=
=
=
=
=
lim
f (x, y) = lim (c + y) = c + a
lim
f (x, y) = lim (6 − y) = 6 − a.
(x,y)→(4,a)
x<4
Correct Answers:
10
0
6
-4
6
-4
y→a
and
(x,y)→(4,a)
x>4
y→a
So we need to see if we can find one value for c such that
c + a = 5 − a for all a. This would require that c = 5 − 2a, but
then c would depend on a, which is exactly what we don’t want.
Therefore, we cannot make the function continuous everywhere.
Graphically we can see this by thinking about what the function looks like: for x ≤ 4, it is a plane with positive y-slope,
passing through the line y = 0, z = c. For x > 4, it is a plane
with negative y-slope, pssing through the line y = 0, z = y0 = 6.
2. (0 pts) Library/Michigan/Chap12Sec5/Q19.pg
Find a function f (x, y, z) whose level surface f = 1 is the graph
of the function g(x, y) = x + 6y.
f (x, y, z) =
SOLUTION
The graph of g(x, y) = x + 6y is the set of all points (x, y, z)
satisfying z = x + 6y, or x + 6y − z = 0. This is a level surface,
but we want the surface equal to the constant value 1, not 0,
Correct Answers:
• none
1
(b) Any vector perpendicular to ~n is parallel to the plane, so
one possible answer is
4. (0 pts) Library/Michigan/Chap12Sec6/Q15.pg
Show that the function
~v = 6~i − 8~j.
x4 y
.
f (x, y) = 8
x + y4
Many other answers are possible.
Correct Answers:
does not have a limit at (0, 0) by examining the following limits.
(a) Find the limit of f as (x, y) → (0, 0) along the line y = x.
lim f (x, y) =
• 8i+6j-k
• 6i-8j
(x,y)→(0,0)
y=x
6. (0 pts) Library/Michigan/Chap13Sec3/Q31.pg
Compute the angle between the vectors − ĩ − j̃ + k̃ and − ĩ + j̃ +
k̃.
radians
angle =
(Give your answer in radians, not degrees.)
SOLUTION
(− ĩ− j̃+ k̃)·(− ĩ+ j̃+ k̃)
cos θ = ||−
= 13 .
ĩ− j̃+ k̃||||− ĩ+ j̃+ k̃||
(b) Find the limit of f as (x, y) → (0, 0) along the line y = x4 .
lim f (x, y) =
(x,y)→(0,0)
y=x4
(Be sure that you are able to explain why the results in (a)
and (b) indicate that f does not have a limit at (0,0)!
SOLUTION
(a) Let us suppose that (x, y) approaches (0, 0) along the line
y = x. Then
f (x, y) = f (x, x) =
Therefore
lim
(x,y)→(0,0)
y=x
x5
x8 + x4
=
So, θ = arccos( 13 ) ≈ 1.23 radians.
Correct Answers:
x
.
x4 + 1
• arccos(1/3)
7. (0 pts) Library/CU/Michigan/Chap13Sec4/Q28.pg
Suppose~v ·~w = 7 and ||~v ×~w|| = 3, and the angle between~v and
~w is θ (measured in radians). Find
(a) tan θ =
(b) θ =
SOLUTION
(a) Since ~v · ~w = ||~v|| ||~w|| cos θ and ||~v × ~w|| = ||~v|| ||~w|| sin θ,
x
f (x, y) = lim 4
= 0.
x→0 x + 1
(b) On the other hand, if (x, y) approaches 99 990,0) the curve
y = x4 we have
f (x, y) = f (x, x4 ) =
x8
x8 + x16
=1
tan θ =
and so
lim
(x,y)→(0,0)
y=x4
f (x, y) = lim f (x, x4 ) = 1.
(b) Then θ = tan−1 (0.428571) = 0.404892.
x→0
Correct Answers:
• 3/7
• arctan(3/7)
Thus no matter how close they are to the origin, there will be
points (x, y) such that f (x, y) is close to 0 and points (x, y) such
that f (x, y) is close to 1. So the limit
lim
||~v × ~w|| 3
sin θ
=
= = 0.428571.
cos θ
~v · ~w
7
8. (0 pts) Library/Michigan/Chap13Sec4/Q13.pg
Find an equation for the plane through the points
(2, 4, 4), (−2, −1, 0), (−2, 0, −1).
The plane is
SOLUTION
The displacement vector from (2, 4, 4) to (−2, −1, 0) is:
f (x, y)
(x,y)→(0,0)
does not exist.
Correct Answers:
• 0
• 1
~a = −4~i − 5 ~j − 4~k.
5. (0 pts) Library/Michigan/Chap13Sec3/Q17.pg
(a) Find a vector ~n perpendicular to the plane
The displacement vector from (2, 4, 4) to (−2, 0, −1) is:
~b = −4~i − 4 ~j − 5~k.
z = 8x + 6y.
Therefore a vector normal to the plane is:
~n =
(b) Find a vector ~v parallel to the plane.
~v =
SOLUTION
(a) Writing the plane in the form 8x + 6y − z = 0 shows that
a normal vector is
~n = 8~i + 6~j −~k.
~n = ~a ×~b = 9~i − 4 ~j − 4~k.
Using the first point, the equation of the plane can be written as:
9x − 4y − 4z = −14.
Correct Answers:
• 9*x-4*y-4*z = -14
Any multiple of this vector is also a correct answer.
2
11. (0 pts) Library/Michigan/Chap14Sec6/Q08.pg
9. (0 pts) Library/Michigan/Chap14Sec3/Q03.pg
Find the equation of the tangent plane to
If
z = (x + 4y) ey ,
x = u,
y = ln(v) ,
find ∂z/∂u and ∂z/∂v. The variables are restricted to domains
on which the functions are defined.
∂z/∂u =
∂z/∂v =
SOLUTION
Since z is a function of two variables x and y which are functions of two variables the two chain rule identities which apply
are:
∂z ∂x ∂z ∂y
∂z
=
+
∂u ∂x ∂u ∂y ∂u
so
∂z
= (ey ) (1) + (4ey + (x + 4y) ey ln(e)) · 0
∂u
= eln(v) · 1.
and
∂z ∂x ∂z ∂y
∂z
=
+
∂v ∂x ∂v ∂y ∂v
so
∂z
1
= (ey )(0) + (4ey + (x + 4y) ey ln(e)) · ( )
∂v
v
1
= 4eln(v) + (u + 4 ln(v)) eln(v) ln(e) .
v
z = ex + y + y3 + 9
at the point (0, 4, 78).
z=
SOLUTION
We have
z = ex + y + y3 + 9.
The partial derivatives are
∂z x
= 1,
=
e
ln(e)
∂x (x,y)=(0,4)
(x,y)=(0,4)
and
∂z 2
= 49.
= 1 + 3y ∂y (x,y)=(0,4)
(x,y)=(0,4)
So the equation of the tangent plane is
z = 78 + 1(x − 0) + 49(y − 4) = 78 + x + 49(y − 4) .
Correct Answers:
• 78+x+49*(y-4)
10. (0 pts) Library/Michigan/Chap14Sec4/Q69.pg
(a) What is the rate of change of f (x, y) = 3xy + y2 at the point
(4, 4) in the direction ~v = 2i + 2 j?
f~v =
(b) What is the direction of maximum rate of change of f at
(4, 4)?
direction =
(Give your answer as a vector.)
(c) What is the maximum rate of change?
maximum rate of change =
SOLUTION
We see that
∇ f = 3yi + (3x + 2y) j,
Correct Answers:
• ((eˆ[ln(v)]))*((1))
• ((4*eˆ[ln(v)]+[u+4*ln(v)]*eˆ[ln(v)]*ln(e)))*((1/v))
12. (0 pts) Library/Michigan/Chap15Sec1/Q07.pg
Find the critical points for the function
f (x, y) = x2 − 2xy + 3y2 − 12y
and classify each as a local maximum, local minimum, saddle
point, or none of these.
critical points:
(give your points as a comma separated list of (x,y) coordinates.)
classifications:
(give your answers in a comma separated list, specifying maximum, minimum, saddle point, or none for each, in the same
order as you entered your critical points)
SOLUTION
To find the critical points, we solve fx = 0 and fy = 0 for x
and y. These equations are
so at the point (4, 4), we have
∇ f = 12i + 20 j.
(a) The directional derivative is
∇ f (4, 4) ·
~v
2i + 2 j
64
= (12i + 20 j) · √
=√ .
|~v|
8
8
(b) The direction of maximum rate of change is
∇ f (4, 4) = 12i + 20 j.
fx = 2x − 2y = 0,
(c) The maximum rate of change is
√
|∇ f (4, 4)| = 544.
and
fy = −2x + 6y − 12 = 0.
We see from the first equation that x = y. Substituting this into
the second equation shows that y = 3. The only critical point is
(3, 3).
We then have
Correct Answers:
• 22.6274
• 12i+20j
• 23.3238
D = ( fxx )( fyy ) − ( fxy )2 = (2)(6) − (−2)2 = 8.
3
Since D > 0 and fxx = 2 > 0, the function f has a local minimum
at the point (3, 3).
or
m(λ) = −17λ2 + 15λ.
(b) The maximum value of m(λ) = −17λ2 + 15λ occurs at
a critical point, where m0 (λ) = −34λ + 15 = 0. At this point,
λ = 0.441176 and m(0.441176) = 3.30882.
(c) We want to minimize f (x, y) = x2 + y2 subject to the constraint g(x, y) = 15, where g(x, y) = 2x + 8y. The method of
Lagrange multipliers has us solve
Correct Answers:
• (3,3)
• minimum
13. (0 pts) Library/Michigan/Chap15Sec2/Q13.pg
Does the function
x2
+ 8y3 + 8y2 − 5x
2
have a global maximum and global minimum? If it does, identify the value of the maximum and minimum. If it does not, be
sure that you are able to explain why.
Global maximum?
(Enter the value of the global maximum, or none if there is no
global maximum.)
Global minimum?
(Enter the value of the global minimum, or none if there is no
global minimum.)
SOLUTION
Suppose x is fixed. Then for large values of y the sign of
f is determined by the highest power of y, namely y3 . Thus,
f (x, y) → ∞ as y → ∞, and there can be no global maximum.
For y → −∞ we similarly have f (x, y) → −∞, so that there is no
global minimum either.
2x = 2λ,
f (x, y) =
2y = 8λ,
and
2x + 8y = 15.
We can see that this gives the same solutions for x and y as we
obtained in part (a), and then, plugging in to the third equation,
we have
2λ + 32λ = 15.
Thus we get the same λ as before: λ = 0.441176, so that the
minimum value of f is f (1λ, 4λ), where λ = 0.441176, or
fmin = 3.30882.
Note that the two question have the same answer. This makes
sense, because in the first problem the largest minimum value
will clearly occur when 2x + 8y = 15, which is the constraint in
the third equation.
Correct Answers:
• 15*L - (2*2 + 8*8)*Lˆ2/4
• 2*15/(2*2 + 8*8)
• 15*15/(2*2 + 8*8)
• 15*15/(2*2 + 8*8)
• 2*15/(2*2 + 8*8)
Correct Answers:
• none
• none
14. (0 pts) Library/CU/Michigan/Chap15Sec3/Q43.pg
For each value of λ the function h(x, y) = x2 + y2 − λ(2x + 8y −
15) has a minimum value m(λ).
(a) Find m(λ)
m(λ) =
(Use the letter L for λ in your expression.)
(b) For which value of λ is m(λ) the largest, and what is that
maximum value?
λ=
maximum m(λ) =
(c) Find the minimum value of f (x, y) = x2 + y2 subject to
the constraint 2x + 8y = 15 using the method of Lagrange multipliers and evaluate λ.
minimum f =
λ=
(How are these results related to your result in part (b)?)
SOLUTION
(a) The critical points of h(x, y) occur where
15. (0 pts) Library/Michigan/Chap16Sec2/Q13.pg
Consider the shaded region in the graph below.
R
Write R f dA on this region as an iterated integral:
RbRd
f
d ,
R dA = a c f (x, y) d
where
,
a=
b=
,
c=
, and
d=
.
SOLUTION
This region lies between x = 0 and x = 6 and between the
lines y = 2x and y = 12, and so the iterated integral is
R
hx (x, y) = 2x − 2λ = 0,
Z 6 Z 12
hy (x, y) = 2y − 8λ = 0.
f (x, y) dy dx.
0
The only critical point is (x, y) = (λ, 4λ) and it gives a minimum
value for h(x, y). That minimum value is
2
2x
Alternatively, we could have set up the integral as follows:
Z 12 Z
2
m(λ) = h(λ, 4λ) = λ + 16λ − λ(2λ + 32λ − 15)
0
4
0
y
2
f (x, y) dx dy.
•
•
•
•
Correct Answers:
• y
• x
• 0
• 6
• 2*x
• 12
4
r*f(r*cos(t),r*sin(t))
r
t
17. (0 pts) Library/Michigan/Chap16Sec5/Q09.pg
Evaluate the triple integral of f (x, y, z) = cos(x2 + y2 ) over the
the solid cylinder with height 6 and with base of radius 2 centered on the z axis at z = −3.
Integral =
SOLUTION
We have
16. (0 pts) Library/Michigan/Chap16Sec4/Q07.pg
For each of the following, set up the integral of an arbitrary
function f (x, y) over the region in whichever of rectangular or
polar coordinates is most appropriate. (Use t for θ in your expressions.)
(a) The region
Z −3 Z 2π Z 2
Z
f dV =
−3
W
0
(cos(r2 ) r dr dθ dz.
0
Integrating,
Z −3 Z 2π Z 2
−3
0
(cos(r2 ) r dr dθ dz =
−3
0
1
(sin(4))
2
With a =
,b=
c=
, and
d
=
R R
integral = ab cd
(b) The region
0
2
r=2
Z
·sin(r2 ) dθ dz =
r=0
Z −3
2π dz = 6π(sin(4)).
−3
Correct Answers:
• 6*pi*sin(2*2)
,
,
d
d
18. (0 pts) Library/CU/Michigan/Chap16Sec7/Q19.pg
In this problem we use the change
of variables x = 3s + t,
R
y = s − 2t to compute the integral R (x + y) dA, where R is the
parallelogram formed by (0, 0), (6, 2), (8, −2),
and (2, −4).
.
First find the magnitude of the Jacobian, ∂(x,y)
∂(s,t) =
Then, with a =
,b=
,
c=
, and d =
,
R
With a =
,b=
,
c=
, and
d
=
,
R R
integral = ab cd
d
d
SOLUTION
(a) Since this is a rectangular region, we use Cartesian coordinates. This gives
R (x
Rb Rd
a
c
x
y
(
s+
t+
)dt ds =
= 3s + t
= s − 2t,
we have
f (x, y) dy dx.
2
+ y) dA =
SOLUTION
Given
Z 7Z 5
∂(x, y) ∂(s,t) = 1
(b) Since this is a partially-circular region, we use polar coordinates. This gives
∂x
∂s
∂y
∂s
hence
∂x
∂t
∂y
∂t
3
=
1
1 = | − 7|,
−2 ∂(x, y) ∂(s,t) = 7.
Z π/4 Z 4
f (r cos(θ), r sin(θ)) r dr dθ.
0
Z −3 Z 2π
1
0
We therefore get
Correct Answers:
• 2
• 7
• 1
• 5
• f(x,y)
• y
• x
• 0
• 0.785398
• 0
∂(x, y) dsdt =
((3s + t) + (s − 2t)) ∂(s,t) T
Z
Z
(x + y) dA =
R
Z
T
(4s − t)(7) ds dt =
Z
(7(4s − t)) ds dt,
T
where T is the region in the st-plane corresponding to R.
Now, we need to find T .
Because
x = 3s + t
y = s − 2t
5
−3 Z 2π
−3
0
we can use the boundaries of the region R to find the boundaries
of T . Using the four vertices of the parallelogram R, we the
edges y = 31 x, y = −2 x, y = 13 (x − 2) − 4 and y = −2(x − 6) + 2.
So, from the above transformation, the boundaries are t = 0,
s = 0, t = 2 and s = 2. Therefore
Z 2Z 2
Z
(x + y) dA =
R
0
• -1*-14/11
• 1
20. (0 pts) Library/Michigan/Chap17Sec1/Q47.pg
(a) Find a vector parallel to the line of intersection of the planes
2x + 3y − z = 0 and −2x − 5y − 4z = 7.
~v =
(b) Show that the point (1, −1, −1) lies on both planes. Then
find a vector parametric equation for the line of intersection.
~r(t) =
SOLUTION
(a) Normal vectors to the two planes are
7(4s − t) ds dt = 84.
0
(Alternately, we could of course also write the integral as
R0R0
2 2 7(4s − t) ds dt.)
Correct Answers:
• 3*2 + 1
• 0
~n1 = 2~i + 3~j −~k and ~n2 = −2~i − 5~j − 4~k.
• (2*6 + 2)/(1 + 3*2)
The vector ~n1 ×~n2 is perpendicular to both planes and parallel
• 0
to the line of intersection:
• (2 - 3*-4)/(3*2 + 1)
• 7*(3 + 1)
~i
~k ~j
• 7*(1 - 2)
~n1 ×~n2 = 2
3 −1 = −17 ĩ + 10 j̃ − 4 k̃.
• 0
−2 −5 −4 • (2*6 + 2)*(2 - 3*-4)*(2 + 2 + 3*2 - 3*-4 + 2*(6 + 3*6 - 2 + 3*-4))/(2*(1 + 3*2)ˆ2)
(b) To check that the point (1, −1, −1) lies on the planes,
substitute into each equation.
19. (0 pts) Library/Michigan/Chap16Sec7/Q13.pg
Find a number a so thatR the
change of variables s = x + ay,t = y
R
transforms the integral R dx dy over the parallelogram R in the
xy-plane with vertices (0, 0), (10, 0), (−14, 11), (−4, 11) into an
integral
Z Z ∂(x, y) ds dt
T ∂(s,t)
2x + 3y − z = (2)(1) + (3)(−1) + (−1)(−1) = 0,
and
−2x − 5y − 4z = (−2)(1) + (−5)(−1) + (−4)(−1) = 7.
Thus, the point lies on both planes. Then a vector parametric
equation of the line is
over a rectangle T in the st-plane.
a=
What is ∂(x,y)
∂(s,t) in this case?
∂(x,y) ∂(s,t) =
SOLUTION
Inverting the change of variables gives x = s − at, y = t. The
four edges of R are
11
11
y = 0, y = 11, y = − x, y = − (x − 10).
14
14
The change of variables transforms the edges to
11
11
11
11
55
t = 0,t = 11,t = − s + at,t = − s + at + .
14
14
14
14
7
These are equations for the edges of a rectangle in the st-plane
if the last two are of the form: s = (Constant). This happens
14
when the t terms drop out, or a = 14
11 . With a = 11 the change of
variables gives
Z Z ∂(x, y) ds dt
T ∂(s,t)
over the rectangle
~r(t) = (1 − 17t) i + (10t − 1) j − (1 + 4t) k.
Correct Answers:
• -17i+10j-4k
• (1,-1,-1)+(-17*i+10*j-4*k)*t
21. (0 pts) Library/Michigan/Chap17Sec5/Q27.pg
Find parametric equations for the sphere centered at the origin
and with radius 6. Use the parameters s and t in your answer.
x(s,t) =
,
, and
y(s,t) =
z(s,t) =
, where
≤s≤
and
≤t ≤
.
SOLUTION
We use spherical coordinates with φ = s and θ = t as the two
parameters. Since the radius is 6, we can take
x = 6 cos(t) sin(s) ,
y = 6 sin(t) sin(s) ,
and
with
0 ≤ s ≤ π and
T : 0 ≤ t ≤ 11, 0 ≤ s ≤ 10.
The jacobian ∂(x,y)
∂(s,t) is
∂(x, y) 1 − 14 =
11 = 1.
∂(s,t) 0
1 Correct Answers:
• 6*cos(t)*sin(s)
• 6*sin(t)*sin(s)
• 6*cos(s)
• 0
• 2*pi
• 0
Correct Answers:
6
0 ≤ t ≤ 2π.
z = 6 cos(s) ,
• pi
24.R (0 pts) Library/Michigan/Chap18Sec2/Q13.pg
Find C ~F · d~r for ~F = 5y~i − (sin y)~j on the curve counterclockwise around the unit circle C starting at the point (1, 0).
R
~
C F · d~r =
SOLUTION
The curve C is parameterized by
22. (0 pts) Library/Michigan/Chap17Sec5/Q31.pg
Consider the cone shown below.
~r = cost~i + sint ~j,
for 0 ≤ t ≤ 2π,
so,
~r0 (t) = − sint~i + cost ~j.
Thus,
Z
If the height of the cone is 9 and the base radius is 4, write a
parameterization of the cone in terms of r = s and θ = t.
,
x(s,t) =
, and
y(s,t) =
z(s,t) =
, with
≤s≤
and
≤t ≤
.
SOLUTION
Since the parameterization is specified to be in terms of the
radius r and angle θ, we find x, y and z in terms of the parameters
r = s and θ = t. We have
Z 2π
(−2 sin2 t − sin (sint) cost)dt
2π
5
= (sint cost − t) + cos (sint)
2
0
= −5π
0
Correct Answers:
• -1*5*pi
25. (0 pts) Library/Michigan/Chap18Sec3/Q31.pg
~ = (yexy + 3 cos(3x + y))~i + (xexy +
For the vector field G
~ along the curve C from
cos(3x + y)) ~j, find the line integral of G
the origin along the x-axis to the point (5, 0) and then counterclockwise around
circumference of the circle x2 + y2 = 25
√ the √
to the point (5/ 2, 5/ 2).
R
~
C G · d~r =
SOLUTION
~ = ∇(exy + sin(3x + y)), the line integral can be calSince G
culated using the Fundamental Theorem of Line Integrals:
(5/√2,5/√2)
Z
√
xy
~F · d~r = e + sin(3x + y) = e25/2 + sin(6 2) − 3.
1
z = 9(1 − s),
4
with
0 ≤ s ≤ 4 and 0 ≤ t ≤ 2π.
There are, of course, other parameterizations, but they will not
be in terms of r and θ, as required in this problem.
Correct Answers:
s*cos(t)
s*sin(t)
9 - (9/4)*s
0
4
0
2*pi
c
(0.0)
Correct Answers:
• 5/sqrt(2)
26. (0 pts) Library/CU/Michigan/Chap18Sec4/Q18.pg
Calculate C ((8x + 7y)~i + (4x + 4y) ~j) · d~r where C is the circular path with center (a, b) and radius m, oriented counterclockwise.
Use Green’s Theorem.
R
((8x
+
7y)~i + (4x + 4y) ~j) · d~r =
C
SOLUTION
Green’s theorem gives
R
23. (0 pts) Library/Michigan/Chap18Sec1/Q09.pg
Calculate the line integral of the vector field ~F = 4 ĩ − 3 j̃ along
the line from the point (1, 0) to the point (7, 0).
The line integral =
SOLUTION
Since ~F is a constant vector field and the curve is a line,
R
~
~
~
C F · d~r = F · ∆~r, where ∆~r = 6i. Therefore,
Z
(5 sint~i − sin (sint) ~j) · (− sint~i + cost ~j)dt
0
=
y = s sint,
•
•
•
•
•
•
•
Z 2π
C
x = s cost,
and
~F · d~r =
Z
(((8x+7y)~i+(4x+4y) ~j)·d~r =
R
C
ZZ
~F · d~r = (4 ĩ − 3 j̃) · 6~i = 24.
=
C
Correct Answers:
• (4-7)*pi*mˆ2
• (7 - 1)*4
7
∂
∂
(4x+4y)− (8x+7y)dA
∂x
∂y
−3dA = −3 · Area of R = −3πm2 .
R
Correct Answers:
ZZ
Z 3Z 3
27. (0 pts) Library/Michigan/Chap18Sec4/Q15.pg
Use Green’s Theorem to calculate the circulation of ~F = 2xy~i
around the rectangle 0 ≤ x ≤ 3, 0 ≤ y ≤ 8, oriented counterclockwise.
circulation =
SOLUTION
By Green’s Theorem, with R representing the interior of the
square,
Z
Z Z
∂
∂
~F · d~r =
(0) − (2xy) dA = −2x dA.
∂y
C
R ∂x
R
=
0
~F · d~r =
Z 3Z 8
0
C
−2x dy dx = −72.
0
Flux =
~v · d~A =
S
3
dy
0
Thus,
28. (0 pts) Library/Michigan/Chap19Sec1/Q07.pg
Find the flux of the constant vector field ~v = −2 ĩ + 4 j̃ − 2 k̃
through a square plate of area 9 in the zx-plane oriented in the
positive y-direction.
flux =
SOLUTION
On the surface, d~A = ~j dA, so only the ~j component of~v contributes to the flux:
Z
2
31. (0 pts) Library/Michigan/Chap19Sec2/Q11.pg
Compute the flux of the vector field ~F = 5x~i + 5y~j through the
surface S, which is the part of the surface z = 36 − (x2 + y2 )
above the disk of radius 6 centered at the origin, oriented upward.
flux =
SOLUTION
Writing the surface S as z = f (x, y) = 36 − x2 − y2 , we have
d~A = (− fx~i − fy~j +~k)dx dy. = (2x~i + 2y~j +~k)dx dy.
Correct Answers:
• -2*3ˆ2*8/2
Z
2
Correct Answers:
• -3*3ˆ2*3 + 5*3ˆ3*3/3 + 5*3*3ˆ3/3
Thus the circulation is
Z
0
Z 3
5
(−6x+5x +5y ) dx dy =
−3x + x3 + 5xy2
3
0
3
Z 3
0 =
18 + 15y2 dy = 18y + y3 = 189.
3 0
0
2
Z
~F ·d~A =
Z
S
5(x~i+y~j)·(2x~i+2y~j +~k) dx dy =
R
Z
10(x2 +y2 ) dx dy.
R
Converting to polar coordinates, we have
Z 2π Z 6
flux =
0
0
5 6
10r2 r dr dθ = (2π)( )r4 = 6480π.
2
0
Correct Answers:
• pi*5*1296
(−2 ĩ + 4 j̃ − 2 k̃) · ~j dA = 4 · 9 = 36.
S
32. (0 pts) Library/CU/Michigan/Chap19Sec3/Q09b.pg
Find the surface area of the region S on the plane z = 8x + 6y
such that 0 ≤ x ≤ 15 and 0 ≤ y ≤ 10 by finding a parameterization of the surface and then calculating the surface area.
29. (0 pts) Library/Michigan/Chap19Sec1/Q21.pg
A parameterization is
2
~
Calculate the flux of the vector field ~F = 6~i + x ~j − 4k, through
x(s,t)
=
,
the square of side 4 in the plane y = 5, centered on the y-axis,
y(s,t)
=
,
and
with sides parallel to the x and z axes, and oriented in the posiz(s,t)
=
,
with
tive y-direction.
≤s≤
and
flux =
≤
t
≤
.
SOLUTION
Then, the surface area =
The only contribution to the flux is from the ~j-component,
SOLUTION
and since d~A = ~jdx dz on the square, S, we have
A parameterization of S is
Z 2Z 2
Z 2 3 2
Z
x
16
2
2
dz = x =·4s,= 64
y=
Flux = (6~i+x ~j −4~k)·d~A =
x ~j · ~jdx dz =
. t, z = 8s + 6t, for 0 ≤ s ≤ 15, 0 ≤ t ≤ 10.
3
3
−2 −2
−2 3 −2
S
We compute
Correct Answers:
∂~r ∂~r
• 1*4*2*(2)ˆ3/3
×
= (~i + 8~k) × (~j + 6~k) = −8~i − 6~j +~k,
∂s ∂t
30. (0 pts) Library/Michigan/Chap19Sec2/Q03.pg
so that
R
∂~r ∂~r √
Calculate the flux integral. S (3~i + 5z~k) · d~A where S is z =
× = 101.
∂s ∂t x2 + y2 with 0 ≤ x ≤ 3, 0 ≤ y ≤ 3, oriented upward.
R
~
~
~
Then
S (3i + 5zk) · d A =
Z
Z Z 10 Z 15 √
√
∂~r ∂~r SOLUTION
dA =
Surface
area
=
dA
=
×
101 ds dt = 150 101
2
2
~
Since z = x + y , we have zx = 2x and zy = 2y, so d A =
∂t
S
R ∂s
t=0 s=0
(−2x~i − 2y~j +~k) dx dy. Thus
Correct Answers:
Correct Answers:
• 9*4
Z
S
(3~i+5z~k)·d~A =
Z 3Z 3
0
(3~i+5(x2 +y2 )~k)·(−2x~i−2y~j +~k) dx dy,
0
8
• s
• t
•
•
•
•
•
•
4
= 3 · π(0.005)3 = 1.5708 × 10−6 .
3
8*s + 6*t
0
15
0
10
sqrt(8*8 + 6*6 + 1)*15*10
Correct Answers:
• 3*4*pi*(0.005)ˆ3/3
35. (0 pts) Library/Michigan/Chap20Sec1/Q01.pg
!
−y~i + x~j
Consider div
.
(x2 + y2 )
33. (0 pts) Library/Michigan/Chap19Sec3/Q01.pg
Compute the flux of the vector field ~F = z~k through the parameterized surface S, which is oriented toward the z-axis and given,
for 0 ≤ s ≤ 3, 0 ≤ t ≤ 4, by
x = 5s + 5t,
y = 5s − 5t,
(a) Is this a vector or a scalar? ?
(b)
Calculate
it:
−y~i+x~j
div (x2 +y2 ) =
+
=
SOLUTION
~ ~
−yi+x j
(a) div (x
is a scalar
2 +y2 )
(b) We have
!
−y~i + x~j
2xy
(−2)xy
div
=
+
= 0.
2
2
2
2
2
(x + y )
(x + y )
(x2 + y2 )2
z = s2 + t 2 .
flux =
SOLUTION
Since S is given by
~r(s,t) = (5s + 5t)~i + (5s − 5t)~j + (s2 + t 2 )~k,
we have
∂~r
∂~r
= 5~i + 5~j + 2s~k and
= 5~i − 5~j + 2t~k,
∂s
∂t
and
~ ~j ~k ∂~r ∂~r i
× = 5 5 2s = (10s + 10t)~i + (10s − 10t)~j − 50~k.
∂s ∂t 5 −5 2t Correct Answers:
• scalar
• 2*1*x*y/(xˆ2 + yˆ2)ˆ(1+1)
• -2*1*x*y/(xˆ2 + yˆ2)ˆ(1+1)
• 0
36. (0 pts) Library/Michigan/Chap20Sec2/Q07.pg
Use the Divergence Theorem to calculate the flux of the vector
fields ~F1 = −3z~i + 3x~k and ~F2 = −3z~i + (8 − 3y)~j + 3x~k through
Since the ~i component of this vector is positive for 0 < s < 3,
the surface S given by the sphere of radius a centered at the
0 < t < 4, it points away from the z-axis, and so has the oppoorigin with outwards orientation. Be sure that you are able to
site orientation to the one specified. Thus, we use
explain your answers geometrically.
∂~r ∂~r
d~A = − × ds dt,
With W giving the interior of the sphere,
R
R
∂s ∂t
~
~
dV =
S F1 · d A = W
and so we have
Z
Z 4Z 3
and
R
R
~F2 · d~A = W
dV =
~F ·d~A = −
(s2 +t 2 )~k · (10s + 10t)~i + (10s − 10t)~j − 50~k dsS dt
S
0 0
SOLUTION
s=3
Z 4Z 3
Z 4 3
The divergence of the fields are
s
2
2
2 = 50
(s + t ) ds dt = 50
+ st dt
∂
∂
∂
3
0 0
0
s=0
div ~F1 = (−3z) + (0) + (3x) = 0.
4
Z 4
∂x
∂y
∂z
= 50 (9 + 3t 2 ) dt = 50(9t + t 3 = 50(36 + 64) = 5000.
and
0
0
∂
∂
∂
Correct Answers:
div ~F2 = (−3z) + (8 − 3y) + (3x) = −3.
∂x
∂y
∂z
• 2*5*5*3*4*(3*3 + 4*4)/3
Hence,
Z
Z
34. (0 pts) Library/CU/Michigan/Chap20Sec1/Q17.pg
~F1 · d~A =
0 dV = 0,
~
~
A smooth vector field F has div F(3, 2, 1) = 3. Estimate the flux
S
W
of ~F out of a small sphere of radius 0.005 centered at the point
and
Z
Z
Z
(3, 2, 1).
~F2 · d~A =
−3 dV = −3
dV = −4πa3 .
W
S
W
flux ≈
This makes sense, because, the vector field ~F1 is flowing
SOLUTION
around the y-axis and is therefore always tangent to the sphere,
Since div F(3, 2, 1) is the flux density out of a small region
surrounding the point (3, 2, 1), we have
so that the flux is always zero. ~F2 , however, adds a component
in the y direction that points in for y > 0 and out (for sufficiently
Flux out of small region around (3, 2, 1)
div ~F(3, 2, 1) ≈
large negative y) for y < 0, and therefore results in a non-zero
Volume of region.
flux.
So
Correct Answers:
• 0
Flux out of region ≈ (div ~F(3, 2, 1)) · Volume of region
9
• 0
• -3
• 4*-3*pi*aˆ3/3
The side of S is given by x2 + y2 = 9, and its height is 3.
(a) Give a parametric equation,~r(t) for the rim, C.
~r(t) =
(You must use angle bracket notation and enter a 3-vector),
with
≤t ≤
.
(b)
If
S
is
oriented outward and downward, find
R
curl (−5y~i + 5x~j + 6z~k) · d~A.
S
R
~
~
~
~
S curl (−5yi + 5x j + 6zk) · d A =
SOLUTION
(a) The equation of the rim, C, is x2 + y2 = 9, z = 0. This is
a circle of radius 3 centered on the z-axis, and lying in the plane
z = 2. We can parameterize this as
37. (0 pts) Library/Michigan/Chap20Sec3/Q02.pg
Compute the curl of the vector field ~F = −3zi + 7y j − 4xk.
curl =
SOLUTION
We have
~k
~j
~i
∂
∂
∂
= j̃.
curl(−3zi + 7y j − 4xk) = ∂x
∂y
∂z
−3z
7y
−4x Correct Answers:
~r(t) = 3 cos(t) i − 3 sin(t) j + 3k,
• j
(b) Use Stokes’ Theorem, with C oriented clockwise when
viewed from above:
38. (0 pts) Library/Michigan/Chap20Sec3/Q19.pg
Three small squares, S1 , S2 , and S3 , each with side 0.05 and
centered at the point (4, 8, 1), lie parallel to the xy-, yz- and xzplanes, respectively. The squares are oriented counterclockwise
when viewed from the positive z-, x-, and y-axes, respectively.
~ has circulation around S1 of −0.0075, around
A vector field G
~ at the point
S2 of 0.25, and around S3 of 1.25. Estimate curl G
(4, 8, 1)
~ ≈
curlG
SOLUTION
~ has its component in the x-direction given
The vector curl G
by
~ x≈
(curl G)
Z
Circulation around S1
−0.0075
= −3.
=
Area inside S1
(0.05)2
(−5y~i + 5x~j + 6z~k) · d~r.
C
Z
curl (−5y~i + 5x~j + 6z~k) · d~A =
Z
S
(−5y~i + 5x~j) · d~r
C
= −15 · Length of curve = −15 · 2π3 = −90π.
Correct Answers:
• 3*cos(t)i-3*sin(t)j+3k
• 0
• 2*pi
• -2*pi*5*3ˆ2
0.25
Circulation around S2
=
= 100.
Area inside S2
(0.05)2
Similar reasoning leads to
~ z≈
(curl G)
Z
Since C is horizontal, the~k component does not contribute to the
integral. The remaining vector field, −5y~i + 5x~j, is tangent to
C, of constant magnitude || − 5y~i + 5x~j|| = 15 on C, and points
in the opposite direction to the orientation. Thus
=
Circulation around S3
1.25
=
= 500.
Area inside S3
(0.05)2
curl (−5y~i + 5x~j + 6z~k) · d~A =
S
Circulation around small square around x-axis
Area inside square
~ y≈
(curl G)
0 ≤ t ≤ 2π.
40.R(0 pts) Library/Michigan/Chap20Sec4/Q15.pg
Find C ~F · d~r where C is a circle of radius 3 in the plane
x +y+z = 4, centered at (4, 1, −1) and oriented clockwise when
viewed
from the origin, if ~F = −4z~j + 6y~k
R
~
C F · d~r =
SOLUTION
Since
~k ~j
~i
∂
∂
∂ = 10~i,
curl ~F = ∂x
∂y
∂z 0 −4z 6y writing S for the disk in the plane enclosed by the circle, Stokes’
Theorem gives
and
Thus,
~ ≈ 100 ĩ + 500 j̃ − 3 k̃.
curl G
Correct Answers:
• 100i+500j-3k
Z
39. (0 pts) Library/CU/Michigan/Chap20Sec4/Q17.pg
The figure below open cylindrical can, S, standing on the xyplane. (S has a bottom and sides, but no top.)
C
~F · d~r =
Z
curl ~F · d~A =
S
Z
10~i · d~A.
S
Now d~A = ~n dA, where ~n is the unit vector perpendicular to the
plane, so
1
~n = √ (~i + ~j +~k).
3
Thus
Z
Z
Z
~ ~ ~
10
10
10
~F d~r = 10~i· i +√j + k dA = √
dA = √ ·Area of disk = √ π32 =
C
S
S
3
3
3
3
10
Correct Answers:
• 10*pi*3*3/sqrt(3)
c
Generated by the WeBWorK system WeBWorK
Team, Department of Mathematics, University of Rochester
11